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Question
Prove that:
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Solution
\[\text{LHS} = \frac{\left( 2n + 1 \right)!}{n!} \]
\[ = \frac{\left( 2n + 1 \right)\left( 2n \right)\left( 2n - 1 \right) . . . . \left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)}{n!}\]
\[ = \frac{\left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n + 1 \right) \right]\left[ \left( 2 \right)\left( 4 \right)\left( 6 \right) . . . . . . . . . \left( 2n \right) \right]}{n!} \]
\[ = \frac{2^n \left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n + 1 \right) \right]\left[ \left( 1 \right)\left( 2 \right)\left( 3 \right) . . . . . . . . . \left( n \right) \right]}{n!}\]
\[ = \frac{2^n \left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n + 1 \right) \right]\left[ n! \right]}{n!}\]
\[ = 2^n \left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n + 1 \right) \right] = \text{RHS}\]
\[ \text{Hence, proved} . \]
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